STRUCTURE OF Ti-48 AND V-48
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) ( June 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks , discovered by Gell-Mann and Zweig, I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ”(2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838.68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). In this photo I present the electromagnetic laws governing the nuclear structure, but a student of Einstein (Dr Th. Kalogeropoulos ) criticised my discovery of nuclear force and structure by believing that the nuclear structure is due to the invalid relativity. In fact, here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. Structure of stable Ti-48 with S = 0 Comparing the stable Ti-48 of S=0 with the unstable Ti-44 of S=0 one concludes that the Ti48 has the same two additional deuterons like p21n21of S=+1 and p22n22 of S=-1. (See my STRUCTURE OF Ti-44, Sc-44 AND Ca-44 ). Of course the stability of Ti-48 is due to the four extra neutrons which make more bonds at the positions p21 and p22 able to overcome the pp repulsions of long range. In the following diagram you see that the extra n23(-1/2) at the blank position between p11 and p21 makes the bonds (n23-p11) and n23-p21). That is, in the structure of Ti-48 the p21(+1/2) forms not only the (p21-n9), the (p21-n15) and the ( p21-n21) as in the case of the unstable Ti-44 but also it forms the additional ( p21-n23). That is, it makes four bonds per proton. Under these four pn bonds per proton the p21 does not turn to neutron, because the four bonds per proton overcome the pp repulsions of long range. Similarly the extra n24(+1/2) at the blank position between p1 and p22 makes the (n24-p1) and (n24-p22). That is, here p22(-1/2) makes the four pn bonds as (p22-n3), (p22-n13), (p22-n22), and (p22-n24) able to overcome the pp repulsions of long range. Moreover the n25(-1/2) which fills the blank position between p3 and p15 contributes to the increase of the binding energies of pn bonds. In the same way the n26(+1/2) which fills the blank position between p9 and p14 contributes to the increase of the binding energies of pn bonds. Note that the high symmetry of the for blank positions contributes more to the increase of the binding energies. In other words the stable Ti-48 consists of the structure of Ca-40 with S=0 (from p1 to n20), in which we add the two deuterons like p21n21 with S=+1 and the p22n22 with S=-1 Then for the stability of T-48 we add at the four blank positions of high symmetry the n23(-1/2), the n24(+1/2), the n25(-1/2) and the n26(+1/2) having the total spin S=0. STRUCTURE OF STABLE Ti-48 WITH S = 0 (The nucleons n17, p17, p18, and n18, of the α particle which are in front of p5, n5, n7 and p7 respectively are not shown here. Similarly the nucleons p19, n19, n20 and p20 of the'' α'' particle which are behind the n6, p6, p8, and n8 respectively are not shown here) ' ' ' ' p12..........n12 ' n11..........p11.........n23 sixth horizontal plane of 5 nucleons ' ' n10..........p10..........n21' ' n26..........p9............n9..........p21 fifth horizontal plane of 7 nucleons' ' n14..........p8............n8............p16' ' p14..........n7............p7...........n16 fourth horizontal plane of 8 nucleons' ' p13..........n6............p6............n15' ' n13..........p5...........n5............p15 third horizontal plane of 8 nucleons' ' n22.........p4............n4' ' p22..........n3............p3............n25 second horizontal plane of 7 nucleons' ' n2............p2' ' n24..........p1...........n1 first horizontal plane of 5 nucleons' Structure of unstable V-48 with S = +4 Comparing the V-48 with Ti-48 we see that the additional proton of V-48 is able to lead to a structure with S = +4. Under this condition the p22(-1/2) and n22(-1/2) of the second plane of Ti-48 is moved here to fill the blank positions of the fifth plane. Thus they become p22(+1/2) and n22(+1/2). In the following diagram you see that p22 makes only the three bonds like (p22-n10), (p22-n22) and (p22-n14) unable to overcome the pp repulsions of long range. Note that the additional proton like p23(+1/2) is not shown here, because it is in front of n9. Thus it makes the (p23-n9). It also makes the pn bond with n18 of the α particle which is not shown here because it is in front of p7. Moreover p23 makes a bond with the additional n23(+1/2), which is also not shown here, because it is in front of p9. In other words the three deuterons like p21n21, p22n22 and p23n23 belong to the nucleons of the fifth horizontal plane having positive spins. Thus the three deuterons give a total spin S = +3. Note that in the same plane belongs also the additional n25(+1/2), which is not shown here, because it is behind the p10. Similarly the additional n26(+1/2) is not shown here, because it is in front of p15. To conclude we see that the spin S = +4 is due to the spin S=+3 of the three additional deuterons in which we add the spins of n25(+1/2) and the n26(+1/2). However the three protons of the three additional deuterons like p21, p22, and p23 give unstable pn bonds because they form only three pn bonds per proton. Under this condition the V-48 of S =+4 decays into the Ti-48 of S=0, under the transformation of p23 of V-48 into the n23 of Ti-48. Especially the p23(+1/2) decays to the n23(-1/2) of the structure of Ti48. Also for the stability of Ti-48 the nucleons of V-48 like the n23(+1/2), the p22(+1/2), the n22(+1/2), the n25(+1/2) and the n26(+1/2) turn to the nucleons of V-48 like the n26(+1/2), the p22(-1/2), the n22(-1/2) the n24(+1/2) and the n25(-1/2) respectively. STRUCTURE OF UNSTABLE V-48 WITH S = +4 (The nucleons n17, p17, p18, and n18, of the 'α'' particle which are in front of p5, n5, n7 and p7 respectively are not shown here. Similarly the nucleons p19, n19, n20 and p20 of the 'α''' particle which are behind the n6, p6, p8, and n8 respectively are not shown here. In this structure also are no shown the additional nucleons like p23, n23, n25, and n26 ) ' ' ' ' p12..........n12''' ' n11..........p11 sixth horizontal square of 4 nucleons ' ' p22.........n10..........p10..........n21' ' n22..........p9............n9..........p21 fifth horizontal plane of 8 nucleons' ' n14..........p8............n8............p16' ' p14..........n7............p7...........n16 fourth horizontal plane of 8 nucleons' ' p13..........n6............p6............n15' ' n13..........p5...........n5............p15 third horizontal plane of 8 nucleons' ' p4............n4' ' n3............p3 second horizontal square of 4 nucleons' ' n2............p2' ' p1...........n1 first horizontal square of 4 nucleons' Category:Fundamental physics concepts